The proof of the abc conjecture

The abc conjecture: there are a finite number of c (= a + b), for

c > rad (abc)^ (1+ ε)

Or

c < K ε x rad (abc)^ (1+ ε) for ALL c.

Let a = d +/- d1 = (p^n x dp) +/- d1; dp is the largest prime for d; d1 is the smallest integer for a to have a d.

Example: a = 17 = 18 – 1 = (3^2x 2) -1; 18 = d, d1 = -1, p = 3, n = 2

Doing the same for b and c;

    b = e +/- e1 = (q^m x ep) +/- e1

    c = f +/- f1 = (w^k x fp) +/- f1

The +/- can be replaced with + only while push – into the d1 (or f1, e1) if necessary.

So, the rad (abc) = rad (a) rad (b) rad (c)

For rad (c) >= c, f1 ≠ 0 is the necessary condition.

For c >= rad (abc), f1 = 0 is the necessary condition; that is, c cannot be a prime.

The sufficient condition (SC): rad (abc) = pqw (dp x ep x fp) < c

Some scenarios can be evaluated for this sufficient condition.

Scenario 1: if d1 = e1 = 0 and there is a h1 (a natural number) while 1 < h1 < min {n, m, k}, then SC = true

Scenario 2: if d1 = 0 and there is a h2 (a natural number) while 1 < h2 < min {n, m, k}, then SC = true

Scenario 3: if e1 = 0 and there is a h3 (a natural number) while 1 < h3 < min {n, m, k}, then SC = true

Scenario 4: all other cases (the uncertainty).

All four cases, the SC = true.

For any give c (with f1 = 0, not a prime), there are S1 (meeting scenario 1), S2, S3 and S4.

Let S = S1 + S2 + S3 + S4

Now, {rad (abc) < c} = {for any c (not a prime, f1 = 0), is S finite?}

A cop-out way for the answer is by tossing a coin (selecting an arbitrary c): head = true; tail = false.

Then, the P (S) = {tail (50%), head (50%)} after infinite many tosses.

Now, we can make a cheating weight (ε > 0, a real number) and add it to rad (abc) side as {rad (abc) ^ (1+ ε)}.

With this cheating weight on the rad side as {rad (abc)^ (1+ ε) < C}, the P (S) = {tail (< 50%), head (> 50%)} of each toss.

After N tosses, P (S, N) = {tail (~ 0), head (~100)}; that is, for any c (a real number while a + b = c) there is always a N (ε) for each ε (a real number) to ensure that

{N (ε) x rad (abc)^ (1+ ε) > c}, N (ε) is the number of toss needed for that (ε).

The above process can be proved in four steps: induction (operational) progressive process.

First, making the above simple tossing (selecting an arbitrary c doing the actual search) process into a game, as follow.

Every game consists of T (=10) tosses, which produces (i tails, j heads), T = i + j = 10 in this case.

So, P (SC = true) = j/T, P (SC = false) = i/T, P the possibility of SC

   ΔP = P (SC = false) - P (SC = true),

   If ΔP > 0, abc conjecture is false.

   If ΔP < 0, abc conjecture is true

Let G = 1 when ΔP < 0; G = 0 when ΔP > 0

This game will be repeated N times.

When N = 1, G1 = (0 or 1)

N = 2, G2 = (0 or 1)

…

N = n, Gn = (0 or 1)

Let G’n = (number of 1) – (number of 0); {(number of 1) + (number of 0) = n}

Theorem 1: If G’n > 0 for all n > N (ε), (N (ε) a large number > 0), then abc conjecture is true.

Second, the cheating: a cheating weight ε is added on one side of the tossing coin.

That is:    ΔX = {rad (abc)^ (1+ ε) - rad (abc)} = rad (abc)^ε

First principle (the indeterminacy): when ΔX = 0, the average of ΔP = 0 after n games (n x T tosses) when n is a large number.

Theorem 2: when ΔX > 0, the average of ΔP < 0 after n games (n x T tosses) when n is large. (This can be proved by actual calculation and search with a finite n).

Third, the induction proof of theorem 1 and 2: in the above, we have proved n = 1 and n = n. Now, by proofing that n = n + 1 is true, the theorem is proved with induction.

Fourth, going beyond the induction: is there a math ghost rascal which can sabotage the above induction proof?

The answer is no: a cheating game cannot be sabotaged even by a ghost rascal; see the ghost rascal conjecture.

      Ghost-rascal conjecture --- For a coin flipping (tossing) game (head vs tail), T is the number times flip as one ‘game’, N is the number times that that ‘game’ is played. If T >= 10 and N >= 10^500, then no amount of sabotage from a Ghost can change the outcome of this game.

See, Ghost-rascal conjecture and the Ultimate Reality

http://prebabel.blogspot.com/2014/02/ghost-rascal-conjecture-and-ultimate.html

That is, the induction proof of theorem 2 with a large n cannot be sabotaged by any math ghost rascal.

With this ghost-rascal guarantee, there is always an N (ε) for each ε (a real number) to ensure that {N (ε) x rad (abc) ^ (1+ ε) > c} for ALL c (= a + b), N (ε) is the number of toss needed for that (ε).

The abc conjecture is now proved.

But what does this abc conjecture mean in the number (or physics) system?

Equation of Wonder: bigger the ΔX, smaller the ΔP < 0.

For every c (= a + b)

Let a = d +/- d1 = (p^n x dp) +/- d1; dp is the largest prime for d; d1 is the smallest integer for a to have a d.

      b = e +/- e1 = (q^m x ep) +/- e1

      c = f +/- f1 = (w^k x fp) +/- f1

Then {p, q, w, dp, ep, fp, d1, e1, f1, n, m, k} are the players for the dynamics of rad (abc).

 Let Q be the dynamics of rad (abc) on those players.

With ΔX (on rad (abc)), there will be a ΔQ.

Corollary 1: ΔQ = | h/ ΔP |; the larger |ΔP < 0| is, the stronger the possibility that abc conjecture is true. That is, the larger |ΔP < 0| is, the smaller ΔQ is.

Now, the equation of wonder can be rewritten as:

        ΔQ = h/ ΔX or (ΔQ x ΔX = h), h is a real number and should be a constant.

|ΔP < 0| = h/ ΔQ is the possibility of whether there is infinite SC {sufficient condition (SC): rad (abc) = pqw (dp x ep x fp) < c} for an arbitrary c (= a + b).

That is, |ΔP < 0| = h/ ΔQ really defines the internal radical/prime dynamics for SC?

The equation {ΔQ x ΔX = h} shows that ΔP (internal radical/prime dynamics) is confined by ΔX (the cheating weight). 

More info about this Equation of Wonder, see the derivation of physics uncertainty equation via the number system at {Multiverse bubbles are now all burst by the math of Nature, http://prebabel.blogspot.com/2013/10/multiverse-bubbles-are-now-all-burst-by.html }.